A) \[C{{H}_{3}}OH\]
B) \[{{C}_{2}}{{H}_{5}}OH\]
C) \[C{{H}_{3}}C{{H}_{2}}COOH\]
D) \[C{{H}_{2}}C{{H}_{2}}CHO\]
Correct Answer: B
Solution :
\[D-\xrightarrow{oxidation}\] aldehyde \[\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}\underset{Acetic\text{ }acid}{\mathop{C{{H}_{3}}COOH}}\,\] Key Idea: \[{{1}^{o}}\] alcohol on oxidation gives aldehyde having same number of carbon and aldehyde on oxidation gives acid having same number of carbon atoms. It means, D will be alcohol having two carbon atoms that is \[{{C}_{2}}{{H}_{5}}OH\] (ethyl alcohol) and the alcohol on oxidation will give \[C{{H}_{3}}CHO\](acetaldehyde) \[\underset{(D)}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{[O]}C{{H}_{3}}CHO\xrightarrow{[O]}C{{H}_{3}}COOH\]
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